Question: $ D = \left[\begin{array}{rrr}3 & -2 & 0 \\ 1 & -2 & 2\end{array}\right]$ $ B = \left[\begin{array}{rr}5 & 3 \\ 2 & 5 \\ -2 & -1\end{array}\right]$ What is $ D B$ ?
Solution: Because $ D$ has dimensions $(2\times3)$ and $ B$ has dimensions $(3\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ D B = \left[\begin{array}{rrr}{3} & {-2} & {0} \\ {1} & {-2} & {2}\end{array}\right] \left[\begin{array}{rr}{5} & \color{#DF0030}{3} \\ {2} & \color{#DF0030}{5} \\ {-2} & \color{#DF0030}{-1}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ D$ , with the corresponding elements in column $j$ of the second matrix, $ B$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ D$ with the first element in ${\text{column }1}$ of $ B$ , then multiply the second element in ${\text{row }1}$ of $ D$ with the second element in ${\text{column }1}$ of $ B$ , and so on. Add the products together. $ \left[\begin{array}{rr}{3}\cdot{5}+{-2}\cdot{2}+{0}\cdot{-2} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ D$ with the corresponding elements in ${\text{column }1}$ of $ B$ and add the products together. $ \left[\begin{array}{rr}{3}\cdot{5}+{-2}\cdot{2}+{0}\cdot{-2} & ? \\ {1}\cdot{5}+{-2}\cdot{2}+{2}\cdot{-2} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ D$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ B$ and add the products together. $ \left[\begin{array}{rr}{3}\cdot{5}+{-2}\cdot{2}+{0}\cdot{-2} & {3}\cdot\color{#DF0030}{3}+{-2}\cdot\color{#DF0030}{5}+{0}\cdot\color{#DF0030}{-1} \\ {1}\cdot{5}+{-2}\cdot{2}+{2}\cdot{-2} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{3}\cdot{5}+{-2}\cdot{2}+{0}\cdot{-2} & {3}\cdot\color{#DF0030}{3}+{-2}\cdot\color{#DF0030}{5}+{0}\cdot\color{#DF0030}{-1} \\ {1}\cdot{5}+{-2}\cdot{2}+{2}\cdot{-2} & {1}\cdot\color{#DF0030}{3}+{-2}\cdot\color{#DF0030}{5}+{2}\cdot\color{#DF0030}{-1}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}11 & -1 \\ -3 & -9\end{array}\right] $